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anonymous
 4 years ago
Find the limit to 0 of....
[(e^2x)1]/x
Is it possible to do this w/out a calculator?
anonymous
 4 years ago
Find the limit to 0 of.... [(e^2x)1]/x Is it possible to do this w/out a calculator?

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zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0do you know l`hopital rule yet?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can't have a zero in the denominater. The answer is 2

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0idk, I'm just practicing for a test and this question was online...

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0you nee to know how to tke a derivative to do this, you will lear that soon

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know how to take a derivative, but how will that help me solve it?

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0take the derivative of the num and the the denom. then run the limit

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0you will have (2e^2x)/1 now run the limi

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, I've got one: Well we know: \[ e^x1\sim x\\ x\to 0 \] So: \[ \lim_{x\to 0}\frac{e^{2x}1}{x}=\lim_{u\to 0}\frac{e^{u}1}{\frac{1}{2}u} \]Where \(u=2x\) Due to that: \[ \lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2 \]
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